Silverware to Gold


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Problem: Black to play and mate in 3.

Hamish Gold of the Otago Chess Club achieved a chess career milestone last month in becoming joint South Island Champion for the first time. The 2010 championship was held in the Hotel Ashburton and attracted a healthy field of 31 players, including four past champions: Peter Fraemohs, Karma (Andy) Machdoem, Arie Nijman (all from Canterbury) and John van Ginkel (Nelson).

Gold's dream tournament started with his defeat of the defending champion Fraemohs in round two. In round four he and van Ginkel lead after both winning their first three games and drew a short game. Gold took the sole lead in the fifth round when his top seeded opponent, Andy Machdoem lost on time in a complex position. Gold held the lead with a draw to Wellington's Ross Jackson, but faltered in the next round against the wily 81 year old veteran Arie Nijman in round seven, miscalculating and losing a piece for insufficient compensation. The final round saw an unusual situation of nine South Island players all having the mathematical chance on becoming South island Champion, with the ineligible Jackson leading by half a point. In the end it was Gold, Machdoem, van Ginkel and Nijman who all won their games to finish tied for first on 6/8 as joint South Island Champion for 2010. As the player with the best tie-break and only first-time winner, Hamish took the trophy back to Dunedin.

Today's game is Gold's must-win game from the final round playing White against Simon Ward of Canterbury.


1.d4 f5
2.e4!? ---
The Staunton Gambit is certainly the most aggressive response to the Dutch Defence, and has positional merit, in that the move 1...f5 for Black does not aid development and weakens f7.

2.--- fxe4
3.f3 exf3
4.Nxf3 g6
Usual here is 4...Nf6, but ...g6 often gets played afterwards. Though Black seems to have an original plan in mind here.
5.Bd3 Bg7
6.Nc3 Nc6!?
The threat on the d-pawn is not real, since 7 0-0 Nxd4? 8 Ng5! Nf6 9 Nxh7! gives White the better game. Kingside development with 6...Nf6 would have transposed back into normal lines, while the move played provokes an advance that requires some care by Black.
7.d5 Ne5
Another choice was 7...Nb4 maintaining pressure on the d5 pawn and threatening to exchange White's light-squared bishop, though after either 8 Bc4 or 8 0-0 White has compensation for the pawn minus.
8.Nxe5 Bxe5
9.Qe2! Bxc3+?
In trading off his one active piece, Black has granted White a dangerous advantage in development, which he converts to a winning attack. Better was 9...d6, when he can meet the attempt to bring the c3 knight into the attack with 9 Ne4 Nf6 10 Nxf6 exf6 11 Bh6 Kf7! and Black has every chance of emerging a pawn ahead.
10.bxc3 Nf6
11.Bh6! ---
Ignoring the threat on the d-pawn, White homes in on the weak squares left by the exchange of the dark-squared bishop and traps the Black king in the centre.
12.--- e6
Black is in serious trouble thanks to his undeveloped queenside and exposed king. While allowing the queen to defend f6 against the threat of 12 Bg7, this move fails to address the situation. It was time to take drastic action with 12...Nxd5!? 13 Qe5 Nf6 14 Bg7 d6 15 Qe2 Bg4 16 Qe3 Kd7 17 Bxh8 Qxh8 and though White holds the advantage after 18 0-0, Black has 2 pawns for the exchange and chances to hold on. Other moves lose quickly.
12.dxe6 dxe6
13.Rd1! ---
The computer prefers the immediate cash-in by 13 Qe5 Rg8 14 Bg5 Ng4 15 Qa5 Qd5 16 Qa4+ Bd7 17 Qxg4 winning a piece. But I like the calmness of Gold's move – allowing Black fewer options to get active and letting the material take care of itself, much as he has played the whole game.
13.--- Qe7
14.Qe5! Rg8
Unpinning, else 15 0-0 will win the knight, but White substitutes one pin for another.
15.Bg5 Nd7
Again if 15...Rf8 16 0-0 wins the knight, while 15...Ng4 meets with 16 Qa5 b6 17 Qa4+ Bd7 18 Qxg4.
16.Qe3 Qf7
Black can't avoid material loss. 16...Nf6 simply loses the knight to 17 0-0, while 16... Qd6 also meets with 17 Rf1! (better than 17 Bxg6+) when the threat of crashing through on e6 wins the black queen and then some.
17.Rf1 Qxf1+
18.Bxf1 Nb6
Allowing mate in three.
19.Rd8+ Resigns

1-0

Solution: 1 ... f5+! 2 gxf6 (2 Kh4 Qh1#) 2... Qf5+ 3 Kh4 Qh5#.